The power of \(x\) which has the greatest coefficient in the expansion of \((1 + \frac{1}{2} x)^{10}\) is
\(x^2\),
\(x^3\),
\(x^5\),
\(x^{10}\).
The coefficient of \(x^k\), where \(1 \leq k \leq 10\) is a whole number, is \[c_k ={^{10}C_k}\left(\frac{1}{2}\right)^k = \binom{10}{k}\left(\frac{1}{2}\right)^k=\frac{10!}{k!(10 - k)!} \left(\frac{1}{2}\right)^k.\]
- For \(k = 2\), this is \(11.25\),
- for \(k = 3\), this is \(15\),
- for \(k = 5\), this is \(7.875\),
- for \(k = 10\), this is \(9.76... \times 10^{-4}\), and so the answer is (b).
Alternatively, look at \(\dfrac{c_{k+1}}{c_k}\). We have
\[\frac{c_{k+1}}{c_k} = \frac{10!}{(k+1)!(9-k)!} \times \frac{k!(10-k)!}{10!} \times \frac{2^k}{2^{k+1}} = \frac{10-k}{2(k+1)}.\]
So the \(c_k\) are growing when \(\dfrac{c_{k+1}}{c_k} > 1\), that is, when \(10-k > 2(k+1) \iff k < \frac{8}{3}\), or \(k \leq 2\).
So if \(k = 2\), then \(\dfrac{c_{k+1}}{c_k} > 1\), which means that \(c_3 > c_2\).
For \(k > 2\), we know \(\dfrac{c_{k+1}}{c_k} < 1\), and so \(c_k\) decreases from its value at \(c_3\).
Therefore \(x^3\) has the greatest coefficient, and the answer is (b).
